Problem: You have found the following ages (in years) of all 4 tigers at your local zoo: $ 22,\enspace 13,\enspace 18,\enspace 16$ What is the average age of the tigers at your zoo? What is the variance? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 4 tigers at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $4$ ages and divide by $4$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{22 + 13 + 18 + 16}{{4}} = {17.3\text{ years old}} $ Find the squared deviations from the mean for each tiger. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $22$ years $4.7$ years $22.09$ years $^2$ $13$ years $-4.3$ years $18.49$ years $^2$ $18$ years $0.7$ years $0.49$ years $^2$ $16$ years $-1.3$ years $1.69$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{22.09} + {18.49} + {0.49} + {1.69}} {{4}} $ $ {\sigma^2} = \dfrac{{42.76}}{{4}} = {10.69\text{ years}^2} $ The average tiger at the zoo is 17.3 years old. The population variance is 10.69 years $^2$.